Question

What is the concentration of `"AlF"_6^"3-"` after mixing `"25 cm"^3` of 0.1 mol/L aluminium ion with `"25 cm"^3` of 1 mol/L fluoride ion?

For `"AlF"_6^"3-", K_text(eq) = 1.0 × 10^25`

Answer 1

I get `["AlF"_6^"3-"] ="0.0050 mol/dm"^3`.

Explanation:

The equilibrium is

`"Al"^"3+" + "6F"^"-" ⇌ "AlF"_6^(3-)`

`K_"eq" = 1.0 × 10^25`.

That means that the reaction will essentially go to completion.

This becomes a limiting reactant problem.

`"Moles of Al"^(3+) = 0.025 color(red)(cancel(color(black)("dm"^3 "Al"^(3+)))) × ("0.010 mol Al"^(3+))/(1 color(red)(cancel(color(black)("dm"^3 "Al"^(3+))))) = "0.000 25 mol Al"^"3+"`

`"Moles of AlF"_6^"3-" color(white)(l)"from Al"^(3+) = "0.000 25" color(red)(cancel(color(black)("mol Al"^(3+)))) × ("1 mol AlF"_6^"3-")/(1 color(red)(cancel(color(black)("mol Al"^(3+))))) = "0.000 25 mol AlF"_6^"3-"`

`"Moles of F"^"-" = 0.025 color(red)(cancel(color(black)("dm"^3 "F"^"-"))) × ("0.10 mol F"^"-")/(1 color(red)(cancel(color(black)("dm"^3 "F"^"-")))) = "0.0025 mol F"^"-"`

`"Moles of AlF"_6^"3-" color(white)(l)"from F"^"-" = 0.0025 color(red)(cancel(color(black)("mol F"^"-"))) × ("1 mol AlF"_6^"3-")/(6 color(red)(cancel(color(black)("mol F"^"-")))) = "0.000 433 mol AlF"_6^(3-)`

`"Al"^(3+)` gives the fewest moles of `"AlF"_6^"3-"`, so `"Al"^(3+)` is the limiting reactant.

Thus, when the reaction is complete, we have `"0.000 25 mol AlF"_6^"3-"` in `"50 cm"^3` of solution.

∴ `["AlF"_6^"3-"] ="0.000 25 mol"/("0.050 dm"^3) = "0.0050 mol/dm"^3`